Merge K Sorted Lists Using Min Heap in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of merging 3 sorted lists using min heap

What is the printed merged list after running the following C++ code that merges 3 sorted linked lists using a min heap?

DSA C++

#include <iostream>
#include <vector>
#include <queue>

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

struct compare {
    bool operator()(ListNode* a, ListNode* b) {
        return a->val > b->val;
    }
};

ListNode* mergeKLists(std::vector<ListNode*>& lists) {
    std::priority_queue<ListNode*, std::vector<ListNode*>, compare> minHeap;
    for (auto node : lists) {
        if (node) minHeap.push(node);
    }
    ListNode dummy(0);
    ListNode* tail = &dummy;
    while (!minHeap.empty()) {
        ListNode* top = minHeap.top();
        minHeap.pop();
        tail->next = top;
        tail = tail->next;
        if (top->next) minHeap.push(top->next);
    }
    return dummy.next;
}

int main() {
    ListNode a1(1), a2(4), a3(5);
    a1.next = &a2; a2.next = &a3;
    ListNode b1(1), b2(3), b3(4);
    b1.next = &b2; b2.next = &b3;
    ListNode c1(2), c2(6);
    c1.next = &c2;
    std::vector<ListNode*> lists = {&a1, &b1, &c1};
    ListNode* merged = mergeKLists(lists);
    while (merged) {
        std::cout << merged->val << " -> ";
        merged = merged->next;
    }
    std::cout << "null" << std::endl;
    return 0;
}

A 1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> null

B 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> null

C 1 -> 1 -> 3 -> 4 -> 5 -> 6 -> null

D 1 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null

💡 Hint

Trace the min heap operations and how nodes are popped and pushed.

The min heap always pops the smallest current node among all lists. The merged list contains all nodes in sorted order including duplicates.

🧠 Conceptual

intermediate

Time complexity of merging K sorted lists using min heap

What is the time complexity of merging K sorted lists with a total of N elements using a min heap?

A O(K log N)

B O(N log K)

C O(N K)

D O(N log N)

💡 Hint

Consider how many times elements are pushed and popped from the heap and the heap size.

Each of the N elements is pushed and popped once from a heap of size at most K, so each heap operation costs O(log K). Total is O(N log K).

🔧 Debug

advanced

Identify the runtime error in min heap merge code

What runtime error will occur when running this code snippet that merges K sorted lists using a min heap?

DSA C++

#include <queue>
#include <vector>

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

struct compare {
    bool operator()(ListNode* a, ListNode* b) {
        return a->val < b->val; // Note: reversed comparison
    }
};

ListNode* mergeKLists(std::vector<ListNode*>& lists) {
    std::priority_queue<ListNode*, std::vector<ListNode*>, compare> minHeap;
    for (auto node : lists) {
        if (node) minHeap.push(node);
    }
    ListNode dummy(0);
    ListNode* tail = &dummy;
    while (!minHeap.empty()) {
        ListNode* top = minHeap.top();
        minHeap.pop();
        tail->next = top;
        tail = tail->next;
        if (top->next) minHeap.push(top->next);
    }
    return dummy.next;
}

A The merged list will be sorted in descending order, not ascending

B Compilation error due to wrong comparator syntax

C Infinite loop because minHeap never empties

D Segmentation fault due to null pointer dereference

💡 Hint

Check the comparator logic and how it affects the priority queue ordering.

The comparator returns true if a->val < b->val, which makes the priority queue behave like a max heap, so the smallest element is not popped first, resulting in descending order merged list.

📝 Syntax

advanced

Identify the syntax error in min heap comparator

Which option contains the correct syntax for a comparator struct to use in a C++ priority_queue for merging K sorted lists?

A struct compare { bool operator()(ListNode a, ListNode b) { return a.val > b.val; } };

B struct compare { bool operator(ListNode* a, ListNode* b) { return a->val > b->val; } };

C struct compare { bool operator()(ListNode* a, ListNode* b) { return a->val > b->val; } };

D struct compare { bool operator[](ListNode* a, ListNode* b) { return a->val > b->val; } };

💡 Hint

Look for the correct operator overloading syntax for function call operator.

The comparator must overload operator() with pointer parameters. Option C is correct syntax. B misses parentheses after operator, C uses objects not pointers, D uses operator[] which is invalid here.

🚀 Application

expert

Number of nodes in merged list after merging K sorted lists

Given 4 sorted linked lists with lengths 3, 5, 2, and 4 respectively, how many nodes will the merged linked list contain after merging all using a min heap?

A 9

B 12

C 10

D 14

💡 Hint

Sum the lengths of all input lists.

The merged list contains all nodes from all lists, so total nodes = 3 + 5 + 2 + 4 = 14.

Practice

1. What is the main advantage of using a min heap to merge K sorted lists?

easy

A. It efficiently finds the smallest element among the heads of all lists.

B. It sorts each list individually before merging.

C. It uses extra memory to store all elements at once.

D. It merges lists by concatenating them without sorting.

Solution

1. Step 1: Understand the role of min heap in merging

   A min heap helps quickly find the smallest element among multiple sorted lists by always keeping the smallest element at the top.

2. Step 2: Compare with other options

   Options A, B, and C do not describe the min heap's advantage correctly. The min heap does not sort lists individually or concatenate without sorting.

3. Final Answer:

   It efficiently finds the smallest element among the heads of all lists. -> Option A

4. Quick Check:

   Min heap = efficient smallest element retrieval [OK]

Hint: Min heap always gives smallest element fast [OK]

Common Mistakes:

• Thinking min heap sorts entire lists
• Assuming min heap stores all elements at once
• Confusing min heap with simple concatenation

2. Which of the following is the correct way to push an element into a min heap in C++ using STL?

easy

A. std::priority_queue minHeap; minHeap.push(5);

B. std::priority_queue<int> minHeap; minHeap.push(5);

C. std::priority_queue minHeap; minHeap.push(5);

D. std::priority_queue minHeap; minHeap.push(5);

Solution

1. Step 1: Identify min heap declaration syntax

   In C++, to create a min heap, use std::priority_queue with std::greater comparator and std::vector as container.

2. Step 2: Check each option

   std::priority_queue minHeap; minHeap.push(5); uses std::greater and vector correctly. std::priority_queue<int> minHeap; minHeap.push(5); is max heap by default. std::priority_queue minHeap; minHeap.push(5); uses std::less which is max heap. std::priority_queue minHeap; minHeap.push(5); uses std::list which is invalid container for priority_queue.

3. Final Answer:

   std::priority_queue minHeap; minHeap.push(5); -> Option C

4. Quick Check:

   Min heap = priority_queue + greater + vector [OK]

Hint: Min heap needs std::greater and vector container [OK]

Common Mistakes:

• Using default priority_queue (max heap)
• Using std::less instead of std::greater
• Using invalid container like std::list

3. What will be the output of the following code snippet that merges 3 sorted lists using a min heap?

#include <iostream>
#include <vector>
#include <queue>

struct Node {
    int val;
    int listIndex;
    int elementIndex;
    Node(int v, int li, int ei) : val(v), listIndex(li), elementIndex(ei) {}
};

struct Compare {
    bool operator()(Node const& a, Node const& b) {
        return a.val > b.val;
    }
};

int main() {
    std::vector lists = {{1,4,5}, {1,3,4}, {2,6}};
    std::priority_queue minHeap;

    for (int i = 0; i < lists.size(); i++) {
        if (!lists[i].empty())
            minHeap.push(Node(lists[i][0], i, 0));
    }

    std::vector<int> merged;
    while (!minHeap.empty()) {
        Node curr = minHeap.top();
        minHeap.pop();
        merged.push_back(curr.val);
        int nextIndex = curr.elementIndex + 1;
        if (nextIndex < lists[curr.listIndex].size()) {
            minHeap.push(Node(lists[curr.listIndex][nextIndex], curr.listIndex, nextIndex));
        }
    }

    for (int val : merged) std::cout << val << " ";
    return 0;
}

medium

A. 1 1 2 3 4 4 5 6

B. 1 2 3 4 4 5 6

C. 1 1 3 4 4 5 6 2

D. 2 1 1 3 4 4 5 6

Solution

1. Step 1: Understand the merging process

   The code pushes the first element of each list into the min heap. Then it repeatedly extracts the smallest element and pushes the next element from the same list until all elements are merged.

2. Step 2: Trace the merged output

   Lists: {1,4,5}, {1,3,4}, {2,6}. Merged order: 1 (from list0), 1 (list1), 2 (list2), 3 (list1), 4 (list0), 4 (list1), 5 (list0), 6 (list2).

3. Final Answer:

   1 1 2 3 4 4 5 6 -> Option A

4. Quick Check:

   Merge sorted lists = 1 1 2 3 4 4 5 6 [OK]

Hint: Push first elements, pop min, push next from same list [OK]

Common Mistakes:

• Mixing order of elements
• Forgetting to push next element after pop
• Ignoring duplicates in merged output

4. Identify the bug in the following code snippet that attempts to merge K sorted lists using a min heap:

std::priority_queue minHeap;
for (auto& list : lists) {
    for (int val : list) {
        minHeap.push(val);
    }
}
std::vector<int> merged;
while (!minHeap.empty()) {
    merged.push_back(minHeap.top());
    minHeap.pop();
}

medium

A. The min heap is declared incorrectly and will cause a compilation error.

B. The code pushes all elements at once, losing the benefit of merging sorted lists efficiently.

C. The code does not pop elements from the min heap, causing an infinite loop.

D. The merged vector is not initialized before use.

Solution

1. Step 1: Analyze the approach used

   The code pushes all elements from all lists into the min heap at once, then pops them to get sorted order.

2. Step 2: Identify inefficiency

   This approach loses the efficiency of merging K sorted lists using min heap by only pushing the first element of each list and pushing next elements on demand.

3. Final Answer:

   The code pushes all elements at once, losing the benefit of merging sorted lists efficiently. -> Option B

4. Quick Check:

   Push all elements at once = inefficient merge [OK]

Hint: Push only first elements, then next after popping [OK]

Common Mistakes:

• Pushing all elements at once
• Ignoring heap declaration correctness
• Forgetting to pop elements in loop

5. Given K sorted linked lists, how can you modify the min heap approach to merge them efficiently if some lists are empty and others have duplicate values?

hard

A. Ignore empty lists and merge only non-empty lists by concatenation without using a min heap.

B. Use a max heap instead of min heap to handle duplicates and empty lists.

C. Push all elements from all lists into the min heap at once, then pop to merge, skipping duplicates manually.

D. Initialize the min heap with the first non-empty element of each list and push next elements only when popping, allowing duplicates naturally.

Solution

1. Step 1: Handle empty lists and duplicates

   Skip empty lists when initializing the min heap. For duplicates, min heap naturally handles them by ordering values.

2. Step 2: Efficient merging strategy

   Push only the first element of each non-empty list into the min heap. When popping, push the next element from the same list if it exists. This keeps memory and operations efficient.

3. Final Answer:

   Initialize the min heap with the first non-empty element of each list and push next elements only when popping, allowing duplicates naturally. -> Option D

4. Quick Check:

   Push first non-empty elements, push next on pop, duplicates allowed [OK]

Hint: Skip empty lists, push next element after pop [OK]

Common Mistakes:

• Pushing all elements at once wasting memory
• Ignoring empty lists causing errors
• Using max heap which breaks order